Can somebody help me on this thanks
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Answer:
a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.
b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.
Step-by-step explanation:
A sequence of Bernoulli trials composes the binomial distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability that a selected component is non-defective is 0.8
This means that
a) Three non-defective components in a batch of seven components.
This is P(X = 3) when n = 7. So
0.0287 = 2.87% probability that three non-defective components in a batch of seven components.
b) 8 non-defective components are drawn before the first defective component is chosen.
Now the order is important, so the we just multiply the probabilities.
8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So
0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.